Ya. Do it using python. Why do you want to use bash when you already have
python?
See, do something like this:-
import os
alldirs = os.listdir("/path/to/dir")
DIRS = [] #Only the dirs you are interested in
for i in alldirs:
if i.find("deploy") is -1: L.append(i)
if i.find("TEMPLATE") is -1: L.append(i)
#Now L contains all the required dirs.
Hope it helps.
On Tue, Oct 26, 2010 at 9:19 PM, Sean Carolan <[email protected]> wrote:
> > Here is the bash one-liner that generates my list,
> > with one name per line:
> >
> > ls -d */ | grep -v -E 'deploy|TEMPLATE' | sed 's/\///'
> >
> > How would you get the output of this into a python list that could
> > then be used in the script? Please forgive my ignorance; I've read
> > through the documentation but am still not clear on this aspect.
>
> Would this be easier to accomplish using os.listdir()?. Basically I
> want to make a list of directories, excluding files and the
> directories or files containing "deploy" and "TEMPLATE".
>
--
Abhijeet Rastogi (shadyabhi)
http://www.google.com/profiles/abhijeet.1989
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